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To solve this problem, we need to calculate the acceleration of point \( B \) on the disk, given that the disk has an angular acceleration \( \alpha = 8 \, \text{rad/s}^2 \) and an angular velocity \( \omega = 3 \, \text{rad/s} \) at the instant shown. Since the disk does not slip at point \( A \), we can proceed with the following approach:

### Step 1: Identify the Components of Acceleration at Point B

For a rotating object, the acceleration of a point on the rim can be broken into two components:
1. **Tangential Acceleration** \( a_t \): Due to angular acceleration \( \alpha \).
2. **Centripetal (or Radial) Acceleration** \( a_r \): Due to angular velocity \( \omega \).

#### Tangential Acceleration \( a_t \)
The tangential acceleration at point \( B \) is given by:
\[
a_t = \alpha \cdot r
\]
where \( r \) is the distance from the center to point \( B \).

In this case:
- \( \alpha = 8 \, \text{rad/s}^2 \)
- \( r = 0.5 \, \text{m} \)

Thus,
\[
a_t = 8 \times 0.5 = 4 \, \text{m/s}^2
\]

#### Centripetal (Radial) Acceleration \( a_r \)
The centripetal acceleration at point \( B \) is given by:
\[
a_r = \omega^2 \cdot r
\]
where \( \omega = 3 \, \text{rad/s} \).

Thus,
\[
a_r = 3^2 \times 0.5 = 9 \times 0.5 = 4.5 \, \text{m/s}^2
\]

### Step 2: Determine the Total Acceleration at Point B

Since point \( B \) lies on the edge of the disk at an angle of \( 45^\circ \) from the horizontal, we need to combine the tangential and radial accelerations as vectors.

- The tangential acceleration \( a_t = 4 \, \text{m/s}^2 \) acts perpendicular to the radius, in the direction of rotation (counterclockwise).
- The radial acceleration \( a_r = 4.5 \, \text{m/s}^2 \) points towards the center of the disk.

To find the resultant acceleration at point \( B \), we calculate the vector sum of these two accelerations.

### Step 3: Calculate the Components of Each Acceleration at \( 45^\circ \)

Since the disk does not slip at \( A \), we assume a fixed orientation relative to \( B \) for simplicity.

The total acceleration at \( B \), \( a_B \), is given by the magnitude:
\[
a_B = \sqrt{a_t^2 + a_r^2}
\]
Substituting values:
\[
a_B = \sqrt{4^2 + 4.5^2} = \sqrt{16 + 20.25} = \sqrt{36.25} \approx 6.02 \, \text{m/s}^2
\]

### Answer
The acceleration of point \( B \) is approximately \( 6.02 \, \text{m/s}^2 \).

Would you like a detailed breakdown of each step or have any questions?

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Here are five questions that relate to this topic:

1. What is the effect of increasing the angular acceleration \( \alpha \) on the tangential acceleration of point \( B \)?
2. How would the acceleration at point \( B \) change if the angular velocity \( \omega \) were doubled?
3. How can we calculate the acceleration of other points on the disk, such as point \( C \), which is at a different angle?
4. What is the significance of the no-slip condition at point \( A \) in determining point \( B \)'s motion?
5. How does the orientation of point \( B \) on the disk affect its tangential and radial acceleration components?

**Tip**: When solving problems involving circular motion, break down accelerations into radial (centripetal) and tangential components to simplify analysis.

The disk has an angular acceleration α = 8 rad/s² and angular velocity ω = 3 rad/s at the instant shown. If it does not slip at A, determine the acceleration of point B.

Calculate the acceleration of point B on a rotating disk with angular acceleration α = 8 rad/s² and angular velocity ω = 3 rad/s. This solution explores tangential and radial accelerations, vector addition, and provides a detailed step-by-step breakdown for engineering students.

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